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Author Topic: Is hot water heavier than cold water, and how can I prove this?  (Read 19088 times)

Offline PmbPhy

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lightarrow - I don't know why you're trying so hard to make this a debate about relativistic mass but I'm really not interested and I won't discuss it.
In your first post of the thread "A poll on relativistic mass" you asked yourself why I believe relativistic mass is merely another name for energy, now you know  :).

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lightarrow
First off you should have put that in the correct thread and not attempted to hijack this thread for that purpose. Second, I asked only who Who here believes that relativistic mass is merely another name for energy, and why?, not whether someone could derive something without appearing to use relativistic mass. So you took that question, took it to another thread and then posted an answer to a question that wasn't asked. I know everything there is about relativistic mass and how to derive just about everything anybody else can in sr so I'm not ignorant of your derivation and attempts to derive things without appearing to use relativistic mass.
 

Offline lightarrow

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lightarrow - I don't know why you're trying so hard to make this a debate about relativistic mass but I'm really not interested and I won't discuss it.
In your first post of the thread "A poll on relativistic mass" you asked yourself why I believe relativistic mass is merely another name for energy, now you know  :).

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lightarrow
First off you should have put that in the correct thread and not attempted to hijack this thread for that purpose. Second, I asked only who Who here believes that relativistic mass is merely another name for energy, and why?, not whether someone could derive something without appearing to use relativistic mass. So you took that question, took it to another thread and then posted an answer to a question that wasn't asked. I know everything there is about relativistic mass and how to derive just about everything anybody else can in sr so I'm not ignorant of your derivation and attempts to derive things without appearing to use relativistic mass.
The explanation that I wrote in my previous post is just one of the reasons, probably that one is just the most ironic (sometimes I am...).
The main reason is that you claimed, in this thread, that it's possible to measure an increase in mass with a mass spectrometer, and this is incorrect and I've already written it a couple of times in this thread.

You could reply that it's not true because relativistic mass is simply E/c2, but this is *my* thesis, not your.
Your thesis is that relativistic mass is not simply E/c2. I have proved that with a mass spectrometer you can measure a relativistic increase of radius of curvature R, from which you can infer an increase in momentum and in energy, but not an increase in mass, unless you *define* mass = E/c2.

I don't understand the meaning of "appearing" in your phrase:

"I'm not ignorant of your derivation and attempts to derive things without appearing to use relativistic mass. "

Where did I use relativistic mass (I mean, a concept of mass which is different from E/c2)?

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lightarrow
« Last Edit: 02/12/2014 13:41:42 by lightarrow »
 

Offline lightarrow

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Quote from: lightarrow
Why didn't you write the equation:

F = ma = mv2/r = qvB

with F, a, v, B in bold carachter?
Because I had no use for it. I wrote down the magnitudes involved rather than the vectors since it was much easier to do so and its exactly correct. One should always make things easier when one can. Didn't you know this about using the equivalent scalar equations from the associated vector equations when possible? If not then I find that surprising. You saw how easy it was doing so. It'd take more work not to. Besides, that's how most physicist derive it, i.e. using the scalar equation and not the vector equation. I take it that you didn't know that either?
I know and I've written many times, probably *most of the times* magnitudes only, or just a component of the vectors, only, even in this forum. But you didn't write if you believe that equation is vectorially valid too.
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Also, you didn't really say why you believe that relativistic mass is merely another name for energy. All you really did is make a baseless assertion that it is merely another name. However it is in fact not merely a different name for it. There are many reasons why but the real and most obvious reason is that explained in the journal article On The Meaning of E = mc2 by Mendel Sachs, Int. J. Theo. Phys., 8 (1973) who writes
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... inertia and energy refer to conceptually different features of matter. On the one hand, inertia per se relates to the resistance to the change of state of motion of a body, as would be caused, in 'particle physics,' by an external force acting on the body (...).
   On the other hand, energy per se is defined in terms of the solutions of the conservation laws.....
Thus we see that energy and mass are conceptually different (though complimentary) features of matter, according to relativity theory.
Several relativists have made this argument, including me.
Ok, then I tell you  :)
Can you *measure* relativistic mass, or some other physical quantity from which you can compute relativistic mass, without measuring momentum p and/or energy E?
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Just because two quantities are proportional it doesn't mean that they are the same thing. E.g. you can't legitimately claim that since the energy is related to the frequency of a photon by E = hf that energy and frequency are the same thing.
Yes, this is true. But it doesn't apply here. You can measure frequency and energy indipendently (actually some physical quantity A from which you compute frequency, some other physical quantity B from which you compute energy, A e B measured independently each other) and *verify* that E = hf holds true and it's not simply a definition. A. Einstein took the Nobel prize even because he showed that E = hf is not simply a definition of (photon) energy knowing the frequency or the other way round.
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Besides, E = mc2 does not work under all situations. E.g. if you have a rod which is at rest lying along the x-axis in the inertial frame S and there is  force on each end of the rod of equal magnitude but opposite directions so as to stress it then it will be no acceleration of the rod but it will have a finite energy. When you transform to S' which is in standard configuration with S and moving in the +x direction then E = mc2 will no longer be true.
I will have to make some computations about it (if I'll be able to do it) because I don't know what to answer about it. How do you define or measure relativistic mass in this case?
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Also, if you are in a non-inertial frame E = mc2 will also not be true. The energy will be proportional to the time component of the momentum 1-form. However relativistic mass will still be the time component of the 4-momentum.
Unfortunately I know almost nothing of GR.
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I didn't want to talk about your mistakes on this subject but since you took this off the thread where I was only looking for opinion you changed the purpose of it.
Mistakes? I can't see where I made them.

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Offline PmbPhy

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lightarrow - I just said that I won't let this thread be hijacked by discussions on relativistic mass, didn't I? And the next thing you did is create consecutive posts about the subject? What's with that?

If you really want to discuss it then you'll either send me a PM and I'll discuss it with you there or create a new thread for the subject.

The worst part about this whole damn thing is that there's too much to cover about all the reasons that went into why relativistic mass became defined and used in relativity, none of which you're mentioned. That's why I wrote that paper on it. So please read it. It's at http://arxiv.org/abs/0709.0687  if you tell me that you've CAREFULLY read it and
http://home.comcast.net/~peter.m.brown/sr/invariant_mass.htm
then I'll take to you about relativistic mass again.
« Last Edit: 02/12/2014 17:05:03 by PmbPhy »
 

Offline phyti39

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I researched this topic a while ago, because its implications are far reaching. What I got from wikipedia is:
at 4°c the water molecules bond in a hexagonal form, which increases the volume per per molecule, thus lowering the density.
If not for this property of water, as mentioned, bodies of water would freeze from the bottom up, eliminating life forms it supports.
« Last Edit: 02/12/2014 21:49:37 by phyti39 »
 

Offline alancalverd

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Close enough to the truth, but isn't it fun watching people trying to outclever each other on a matter which may well be completely irrelevant to the original question!
 

Offline lightarrow

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What I got from wikipedia is:
at 4°c the water molecules bond in a hexagonal form, which increases the volume per per molecule, thus lowering the density.
If not for this property of water, as mentioned, bodies of water would freeze from the bottom up, eliminating life forms it supports.
Ok, but the phrase you have quoted (by the way, which is the link to the wiki page?) could have been expressed better, e.g. something like:
"from 4°C onwards the water molecules bond in a hexagonal form ...thus lowering the liquid density"
otherwise it seems that at exactly 4°C water is less dense than at the other temperatures...

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Offline phyti39

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What I got from wikipedia is:
at 4°c the water molecules bond in a hexagonal form, which increases the volume per per molecule, thus lowering the density.
If not for this property of water, as mentioned, bodies of water would freeze from the bottom up, eliminating life forms it supports.
Ok, but the phrase you have quoted (by the way, which is the link to the wiki page?) could have been expressed better, e.g. something like:
"from 4°C onwards the water molecules bond in a hexagonal form ...thus lowering the liquid density"
otherwise it seems that at exactly 4°C water is less dense than at the other temperatures...

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lightarrow
You seemed to understand the graph with your comment in #20.
I just stated the essential factors in words, adding to what was already posted. If the readers have basic comprehension skills, there is no need to keep reinventing the wheel!
 

Offline PmbPhy

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Quote from: alancalverd
A molecule is not an ion. Which seems a good reason to me.
Since I was referring to an ion and not a molecule this comment is moot. However if you've been laboring under the notion that a molecule has to be neutral to be a molecule then that's a very trivial, misleading and uncalled for minor distinction. In any case you should have mentioned this trivial difference in your objection. In any case I don't know where you got your definition claiming that an ion isn't a molecule but chemistry textbooks don't define molecule in that way. I've only see Wikipedia define it as such.

For example: consider the Encyclopedia Britannica
http://www.britannica.com/EBchecked/topic/388236/molecule
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Molecule, the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance.
If you search that page you'll see nothing about a requirement of a molecule being electrically neutral. In fact it speaks of ions in the same page and thus in the same breath, i.e.
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Not all substances are made up of distinct molecular units. Sodium chloride (common table salt), for example, consists of sodium ions and chloride ions arranged in a lattice so that each sodium ion is surrounded by six equidistant chloride ions and each chloride ion is surrounded by six equidistant sodium ions. The forces acting between any sodium and any adjacent chloride ion are equal. Hence, no distinct aggregate identifiable as a molecule of sodium chloride exists. Consequently, in sodium chloride and in all solids of similar type—in general, all salts—the concept of the chemical molecule has no significance. The formula for such a compound, however, is given as the simplest ratio of the atoms—in the case of sodium chloride, NaCl.

Ion is defined here: http://www.britannica.com/EBchecked/topic/292705/ion
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ion,  any atom or group of atoms that bears one or more positive or negative electrical charges. Positively charged ions are called cations; negatively charged ions, anions. Ions are formed by the addition of electrons to, or the removal of electrons from, neutral atoms or molecules or other ions; by combination of ions with other particles; or by rupture of a covalent bond between two atoms in such a way that both of the electrons of the bond are left in association with one of the formerly bonded atoms. Examples of these processes include the reaction of a sodium atom with a chlorine atom to form a sodium cation and a chloride anion; the addition of a hydrogen cation to an ammonia molecule to form an ammonium cation; and the dissociation of a water molecule to form a hydrogen cation and a hydroxide anion.
So when I referred to the "ion H2O" I was exactly right. You misread it and thus changed the subject in your mind to only neutral molecules for some odd reason with no mention of what "I" was actually talking about.

And you've yet to explain why you changed the subject from "ion H2O" to "water molecules"? Would you care to explain that, please? :)
 

Offline alancalverd

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I apologised way back in reply #43.

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Molecule, the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance.

Ions do not retain the chemical properties of the unionised molecule. Water forms the greater part of living cells and does not attack mitotic DNA, but ionised water does, and this is the mechanism by which ionising radiation induces defects and causes cancer - or possibly evolution. The distinction is really quite important.
 

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